All chords pass through the midpoint of each other

Question

Bob draws $n$ chords in a circle, and every chord passes through the midpoint of another chord.

Is it true that all $n$ chords are diameters?

I am not sure how to do these problems.

I tried drawing some circles and lines and it seems always true but I do not know why.

Answer 1

Assume you have one of them, which has a given length and assume this length is smaller than the diameter. Then it is passing through the middle of another chord in the set. How does the length of this one compare to the first? From this you should be able to conclude.

Answer 2

A good start would be this. Take an arbitrary chord. Then consider a second chord that goes through the middle of the first chord.

Here, $AB$ is an arbitrary chord, $C$ is its midpoint, and $DE$ is a chord that goes through $C$. $F$ is the midpoint of $DE$. What can you tell about $OF$ and $OC$?

Unless $O \equiv F \equiv C$, then $OF < OC$. That's because $OF$ is perpendicular to $FC$.

But that is problematic, because the farthest chord from the center still needs to go through the center of a chord.

Here's a formal proof:

Let $DE$ be the chord furthest away from the center, along the perpendicular line from the center to the chord (if there are several equidistant, it can be any of them). $F$ is the midpoint of $DE$. Assume $DE$ is not a diameter.

There exists a chord $AB$ such that $DE$ goes through the midpoint of $AB$. Call that midpoint $C$. Since DE is furthest away, then $OC \leq OF$. $DE$ is not a diameter, so $O$, $F$, and $C$ don't coincide. But $\angle OFC$ is 90 degrees, so $OF < OC$, contradicting that $DE$ is furthest away.

So $DE$ must be diameter. Since it's the furthest chord from the center, all chords must be diameters.